Forking and Dividing in Random Graphsgconant/Talks/Random_Graphs_Talk_handout.… · Gabriel Conant...

Forking and Dividing in Random Graphs

Gabriel Conant

UIC

Graduate Student Conference in LogicUniversity of Notre Dame

April 28-29, 2012

Gabriel Conant (UIC) Forking and Dividing in Random Graphs April 29, 2012 1 / 27

Definitions

(a) A formula ϕ(x , b) divides over a set A ⊂M if there is a sequence(bl)l<ω and k ∈ Z+ such that bl ≡A b for all i < ω and{ϕ(x , bl) : l < ω} is k -inconsistent.

(b) A partial type divides over A if it proves a formula that divides overA.

(c) A partial type forks over A if it proves a finite disjunction offormulas that divide over A.

(d) Given a theory T and M |= T , we define the ternary relations | d

and | on {c ∈M} × {A ⊂M}2 by

c | dA

B ⇔ tp(c/AB) does not divide over A,

c |A

B ⇔ tp(c/AB) does not fork over A.


Facts

Theorem

A partial type π(x , b) (with π(x , y) a type over A) divides over A if andonly if there is a sequence (bl)l<ω of indiscernibles over A with b0 = band

⋃l<ω π(x , bl) inconsistent.

DefinitionA theory T is simple if there is some formula ϕ(x , y) with the treeproperty.

Theorem

A theory T is simple if and only if for all B ⊂M and p ∈ Sn(B) there issome A ⊆ B with |A| ≤ |T | such that p does not divide over A.


SettingLet L = {R}, where R is a binary relation symbol. We write xRy ,

rather than R(x , y). If A and B are sets, we write ARB to mean thatevery point in A is connected to every point in B; and A 6RB to meanthat no point in A is connected to any point in B.

We write A ⊂M to mean that A is a subset of M with |A| < |M|.

Given A ⊆M, let L0A ⊆ LA be the set of conjunctions of atomic and

negated atomic formulas such that no conjunct is of the form xi = a, forsome variable xi and a ∈ A. When we refer to formulas ϕ(x , y) in L0

A,we will assume no conjunct is of the form xi = xj or yi = yj for distincti , j .

Let LRA be the set of L0

A-formulas ϕ(x , y) containing no conjunct ofthe form xi = yj for any i , j .


The Random GraphWe let T0 be the theory of the random graph, i.e., T0 contains

1 the axioms for nonempty graphs,

∀x¬xRx ∀x∀y(xRy ↔ yRx) ∃x x = x ;

2 “extension axioms” asserting that if A and B are finite disjoint setsof vertices then there is some c such that cRA and c 6RB.

T0 is a complete, ℵ0-categorical theory with quantifier eliminationand no finite models.

We fix an uncountable saturated monster model M |= T0 and definethe ternary relation | ∩ on {c ∈M} × {A ⊂M}2 such that

c | ∩A

B ⇔ c ∩ B ⊆ A.


The Random Graph is Simple (first proof)

Theorem (Kim & Pillay)A theory T is simple if and only if there is a ternary relation | o

satisfying automorphism invariance, extension, local and finitecharacter, symmetry, transitivity, monotonicity, and independence overmodels. Moreover, in this case | o, | d , and | are all the same.

Independence over models: Let M |= T , M ⊆ A,B and A | oM

B.Suppose a, b ∈M such that a ≡M b, a | o

MA, and b | o

MB. Then there

is c ∈M such that tp(a/A) ∪ tp(b/B) ⊆ tp(c/AB) and c | oM

AB.

PropositionThe ternary relation | ∩ satisfies all the necessary properties towitness that T0 is simple. In particular, we have

c |A

B ⇔ c ∩ B ⊆ A.


The Random Graph is Simple (second proof)

Lemma

Let A ⊂M, b, c ∈M such that c | ∩A

b. If (bl)l<ω is A-indiscernible,with b0 = b, then there is d ∈M with d | ∩

A

⋃l<ω bl such that

d bl ≡A cb for all l < ω.

Theorem

Suppose B ⊂M |= T0 and p ∈ Sn(B). Then p does not divide overA = {b ∈ B : ∃i xi = b ∈ p(x)}.

1 A theory T is supersimple if for all p ∈ Sn(B) there is some finiteA ⊆ B such that p does not divide over A. Therefore T0 issupersimple.

2 (Kim) If | ∩ is replaced by | in the above lemma, the statementremains true in any simple theory.


Dividing in the Random Graph

Theorem

Let A ⊂M |= T0, ϕ(x , y) ∈ L0A, and b ∈M such that ϕ(x , b) is

consistent. Then ϕ(x , b) divides over A if and only if ϕ(x , b) B xj = bfor some b ∈ b\A.


The Kn-free Random GraphFor n ≥ 3, we let Tn be the theory of the Kn-free random graph, i.e.,

Tn contains

1 the axioms for nonempty graphs;2 a sentence asserting that the graph is Kn-free;3 “extension axioms” asserting that if A and B are finite disjoint sets

of vertices, and A is Kn−1-free, then there is some c such that cRAand c 6RB.

Tn is a complete, ℵ0-categorical theory with quantifier eliminationand no finite models.


The Kn-free Random Graph is Not Simple (first proof)

Theorem

| d -independence over models fails in Tn.

Proof.b1 b2 b3 bn−1

a1 a2 a3 an−1

M

. . .Claim:If b ∈M\A, b 6RA, l(b) < n − 1, and p isin S1(Ab), then p does not divide over A.

So ai | dM{aj : j < i} and bi | d

Mai .

Clearly, {tp(bi/Mai) : i < n}cannot be amalgamated.


The Kn-free Random Graph is Not Simple (second proof)

DefinitionLet k ≥ 3. A theory T has the k -strong order property, SOPk , if thereis a formula ϕ(x , y), with l(x) = l(y), and (bl)l<ω such that

M |= ϕ(bl , bm) ∀ l < m < ω;

M |= ¬∃x1 . . . ∃xk (ϕ(x1, x2) ∧ . . . ∧ ϕ(xk−1, xk ) ∧ ϕ(xk , x1)).

Fact: . . .⇒ SOPk+1 ⇒ SOPk ⇒ . . .⇒ SOP3 ⇒ tree property.

TheoremFor all n ≥ 3, Tn has SOP3.



Proof (Tn has SOP3).

x1

x2

x3

Km ∼=

Km ∼=

Km ∼=

y1

y2

y3

∼= Km

∼= Km

∼= Km

Assume n 6= 4.Let m = dn

3e. Then 2m < n.

Let ϕ(x1, x2, x3, y1, y2, y3)describe this configuration.

We can construct an infinitechain in M |= Tn.

No 3-cycle is possible.



DefinitionLet k ≥ 3. A theory T has the k -strong order property, SOPk , if thereis a formula ϕ(x , y) and (bl)l<ω such that

M |= ϕ(bl , bm) ∀ l < m < ω;

M |= ¬∃x1 . . . ∃xk (ϕ(x1, x2) ∧ . . . ∧ ϕ(xk−1, xk ) ∧ ϕ(xk , x1)).

Fact: . . .⇒ SOPk+1 ⇒ SOPk ⇒ . . .⇒ SOP3 ⇒ tree property.

TheoremFor all n ≥ 3, Tn has SOP3.

Fact: For all n ≥ 3, Tn does not have SOP4.


Recall: Dividing in the Random Graph

Theorem

Let A ⊂M |= T0, ϕ(x , y) ∈ L0A, and b ∈M such that ϕ(x , b) is

consistent. Then ϕ(x , b) divides over A if and only if ϕ(x , b) B xj = bfor some b ∈ b\A.


Dividing in Tn

DefinitionSuppose A,B ⊂M are disjoint. Then B is n-bound to A if there is

B0 ⊆ A ∪ B, |B0| = n, B0 ∩ A 6= ∅ 6= B0 ∩ B, such that

1 (B0 ∩ A)R(B0 ∩ B),2 (B0 ∩ A) ∼= Km, where m = |B0 ∩ A|.

Informally, B is n-bound to A if there is a subgraph B0 ⊆ AB of size n,such that the only thing preventing B0 from being isomorphic to Kn is apossible lack of edges between points in B.

A A

b1 b2 b1 b2

b is 4-bound to Ab is not 4-bound to A(but A is 4-bound to b)


Dividing in Tn

Theorem

Let A ⊂M, ϕ(x , y) ∈ LRA , and b ∈M\A such that ϕ(x , b) is consistent.

Then ϕ(x , b) divides over A if and only if

1 b is not n-bound to A,2 b is n-bound to Ac for any c |= ϕ(x , b).

Proof.(⇐) : b not n-bound to A allows construction of a sequence (bl)l<ω,with enough edges, so that being n-bound to Ac will force(n − 1)-inconsistency in {ϕ(x , bl) : l < ω}.(⇒): Let (bl)l<ω, indiscernible over A, witness dividing. Since ϕ(x , b)does not divide over A in T0, there must be a copy of Kn in Ac(bl)l<ω,where c is an “optimal” solution of {ϕ(x , bl) : l < ω} in T0. Byindiscernibility, this Kn will “project” to the required conditions in Ab.


Dividing in Tn - Examples

CorollarySuppose A ⊂M and b1, . . . ,bn−1 ∈M\A are distinct. Then the formula

ϕ(x , b) :=n−1∧i=1

xRbi

divides over A if and only if b is not n-bound to A.

Corollary

Let A ⊂M and ϕ(x , y) ∈ LRA . Suppose b ∈M\A such that ϕ(x , b) is

consistent and divides over A. Define

Rϕ = {u ∈ Ab : ∃i ϕ(x , b) B xiRu} ∪ {xi : ∃u ∈ Ab, ϕ(x , b) B xiRu}.

Then |Rϕ| ≥ n and |b ∩ Rϕ| > 1.


Recall: | in T0

TheoremLet A,B, c ⊂M |= T0. Then

c |A

B ⇔ c | dA

B ⇔ c ∩ B ⊆ A.


| in Tn

Lemma

Let A ⊂M |= Tn and p ∈ Sm(A). If p ` x = b for some b ∈M thenb ∈ A. If p ` xRb for some b ∈M, then either b ∈ A or p ` x = a forsome a ∈ A.

TheoremLet A,B, c ⊂M |= Tn. Then

c |A

B ⇔ c | dA

B ⇔ c ∩ B ⊆ A and for all b ∈ B\A, b iseither n-bound to A or not n-bound to Ac.

CorollaryLet A ⊂M |= Tn. If π(x) is a partial type over A, then π(x) does notfork over A.


Forking in Tn

Lemma

Suppose (bl)l<ω is an indiscernible sequence in M |= Tn such thatl(b0) = 4 and b0 is K2-free. Then either there are i < j such that{bl

i ,blj : l < ω} is K2-free, or

⋃l<ω bl is not K3-free.


Forking in Tn

TheoremLet M |= Tn and Ab ⊂M such that l(b) = 4, bRA, A ∼= Kn−3, and b isK2-free. For i 6= j , let

ϕi,j(x ,bi ,bj) = xRbi ∧ xRbj ∧∧a∈A

xRa,

and set ϕ(x , b) :=∨i 6=j

ϕi,j(x ,bi ,bj).

Then ϕ(x , b) forks over A but does not divide over A.

A ∼= Kn−3

b1 b2 b3 b4

ϕ(x , b) = “xRA ∧ |{bi : xRbi}| = 2”


Higher Arity GraphsNow suppose R is a relation of arity r ≥ 2. An r -graph is anL-structure satisfying the following sentence

∀x1 . . . ∀xr

R(x1, . . . , xr )→

∧i 6=j

xi 6= xj ∧∧σ∈Sr

R(xσ(1), . . . , xσ(r))

.

So R can be thought of as a collection of r -element subsets of ther -graph.

Let T r0 and T r

n , for n > r , be the natural analogs of T0 and Tn,respectively, to r -graphs.

Call T r0 the theory of the random r -graph, and T r

n the theory ofthe random K r

n -free r -graph, where K rn is the complete r -graph of

size n.T r

n can be thought of as the theory of the unique (countable)Fraısse limit of the class of finite K r

n -free r -graphs.


Recall - Dividing in T0

Lemma

Let A ⊂M |= T0, b, c ∈M such that c | ∩A

b. If (bl)l<ω isA-indiscernible, with b0 = b, then there is d ∈M with d | ∩

A

⋃l<ω bl

such that d bl ≡A cb for all l < ω.

LemmaLet A ⊂M |= T0 and p(x , y) ∈ S(A) such that xi 6= yj ∈ p(x , y) for alli , j . Suppose p(x , b) is consistent and (bl)l<ω is indiscernible over Awith b0 = b. Then there is a solution c of

⋃l<ω p(x , bl).


Dividing in T r0

LemmaLet A ⊂M |= T r

0 and p(x , y) ∈ S(A) such that xi 6= yj ∈ p(x , y) for alli , j . Suppose p(x , b) is consistent and (bl)l<ω is indiscernible over Awith b0 = b. Then there is a solution c of

⋃l<ω p(x , bl).

In fact, we can take c to be an optimal solution, in particular ifA0 ⊆ A(bl)l<ω, then for any i1, . . . , ik

M |= R(ci1 , . . . , cik ,A0) ⇔ R(xi1 , . . . , xik ,A0) ∈⋃l<ω

p(x , bl).

Corollary (T r0 is simple.)

Suppose B ⊂M |= T r0 and p ∈ S(B). Then p does not fork over

A = {b ∈ B : ∃i xi = b ∈ p(x)}.


Dividing in T rn , r > 2

LemmaAssume r > 2. Let A ⊂M |= T r

n and p(x , y) ∈ S(A) such thatxi 6= yj ∈ p(x , y) for all i , j . Suppose p(x , b) is consistent and (bl)l<ω isindiscernible over A with b0 = b. Then there is a solution c of⋃

l<ω p(x , bl).

Corollary (T rn is simple if r > 2.)

Suppose r > 2 and B ⊂M |= T rn and p ∈ S(B). Then p does not fork

over A = {b ∈ B : ∃i xi = b ∈ p(x)}.


Dividing in T rn , r > 2

Proof of Lemma.

A

b0

b1

b2

b3

b4

...

c ∈M′ |= T r0 is an optimal

solution to⋃

l<ω p(x , bl).

c

Show Ac(bl)l<ω is K rn -free.

Let K rn∼= W ⊆ Ac(bl)l<ω.

W ∼= K rn

There is ci ∈W ∩ c.

ci

If |l : W ∩ bl 6= ∅}| > 1

u

v

pick W0 ⊆W\{ci ,u, v},with |W0| = r − 3.

W0

R(ci ,u, v ,W0) ∈⋃

l<ω p(x , bl).

|{l : W ∩ bl 6= ∅}| ≤ 1

W ∼= K rn

bm

“W ∼= K rn ” ∈ p(c, bm)

This contradicts M |= ∃xp(x , bm).


References

B. Kim & A. Pillay, Simple theories, Annals of Pure and AppliedLogic 88 (1997) 149-164.

D. Marker, Model Theory: An Introduction, Springer, 2002.

K. Tent & M. Ziegler, A Course in Model Theory, Cambridge, 2012.


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